A curve is defined by the parametric equations $x=\sin(t)$ and $y=-\cos(t)+\pi$. What is $\dfrac{d^2y}{dx^2}$ in terms of $t$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\tan(t)\sec(t)$ (Choice B) B $\sec^3(t)$ (Choice C) C $\tan(t)$ (Choice D) D $\cos^3(t)$
We are asked to find the second derivative of a parametric function. Recall that the first derivative of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ is found with the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ Then, the second derivative is found with this following rule: $\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{\dfrac{d}{dt}\left(\dfrac{v'(t)}{u'(t)}\right)}{u'(t)}$ Let's start by finding $\dfrac{dy}{dx}$. $\dfrac{dy}{dx}=\tan(t)$ Now we can find $\dfrac{d^2y}{dx^2}$. $\begin{aligned} \dfrac{d^2y}{dx^2}&=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)} \\\\ &=\dfrac{\dfrac{d}{dt}\left[\tan(t)\right]}{\dfrac{d}{dt}[\sin(t)]} \\\\ &=\dfrac{\sec^2(t)}{\cos(t)} \\\\ &=\sec^2(t)\cdot\dfrac{1}{\cos(t)} \\\\ &=\sec^2(t)\cdot\sec(t) \\\\ &=\sec^3(t) \end{aligned}$ In conclusion, $\dfrac{d^2y}{dx^2}=\sec^3(t)$.